Question

In a marketing survey, a random sample of 998 supermarket shoppers revealed that 276 always stock up on an item when they find that item at a real bargain price.

(a) Let *p* represent the proportion of all supermarket
shoppers who always stock up on an item when they find a real
bargain. Find a point estimate for *p*. (Use 3 decimal
places.)

(b) Find a 95% confidence interval for *p*. (Use 3 decimal
places.)

lower limit | |

upper limit |

(c) What is the margin of error based on a 95% confidence interval?
(Use 3 decimal places.)

Answer #1

Solution :

Given that n = 998, x = 276

(a)

**=> proportion p** = x/n

= 276/998

= 0.2766

**= 0.277** (rounded)

(b)

=> p = 0.277 ; q = 1 - p = 0.723

=> for 95% confidence interval, Z = 1.96

=> A 95% confidence interval of the proportion p is

=> p +/- Z*sqrt(p*q/n)

=> 0.277 +/- 1.96*sqrt(0.277*0.723/998)

=> 0.2492 < p < 0.3048

=> 0.249 < p < 0.305 (rounded)

**=> Lower limit = 0.249**

**=> Upper limit = 0.305**

(c)

**=> Margin of error E** = Z*sqrt(p*q/n)

= 1.96*sqrt(0.277*0.723/998)

= 0.0278

= **0.028** (rounded)

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(a) Let p represent the proportion of all supermarket
shoppers who always stock up on an item when they find a real
bargain. Find a point estimate for p. (Use 3 decimal
places.)
(b) Find a 99% confidence interval for p. (Use 3 decimal
places.)
lower limit
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